Question

A block of ice of mass $50kg$ is sliding on a horizontal plane. It starts with speed $5m/s$ and stops after moving through some distance. The mass of ice that has melted due to friction between the block and the surface is:

[Assuming that no energy is lost and latent heat of fusion of ice is $80cal/g$, $J=4.2J/cal$ ]

• A. $2.86g$
• B. $3.86g$
• C. $0.86g$
• D. $1.86g$

Answer 1

The correct option is D $1.86g$

Latent heat of fusion $L_f=80$ $cal/g$ $=80\times 4.2=336$ $J/g$

Mass of ice block initially $M=50kg$

Initial velocity of ice block $u=5m{s}^{-1}$

Final velocity of ice block $v=0$

From work-energy theorem, $W_{friction}=\Delta K.E$

$\therefore$ $W=\frac{1}{2}M{u}^2=\frac{1}{2}\times 50\times 5^2=625J$

Let mass of ice melted be $m$

$\therefore$ $W=m{L}_f$

or, $625=m\times 336$ $⟹m=1.86g$